Question

Prove that curve(x/a)n+(y/b)n=2 touches the straight line x/a+y/b=2 at (a,b) for all values of n belongs to N at the point (a,b)..

Answer 1

The question asked is how to prove that the tangent to the curve (xa)n+(ya)n=2(xa)n+(ya)n=2 at the point (a,b)(a,b) is xa+yb=2?xa+yb=2?

There appears to be a typographical error in the question. If the point (a,b)(a,b) is to lie on this curve, then the equation of the curve should be (xa)n+(yb)n=2.(xa)n+(yb)n=2.

Differentiating the equation of the curve, we get,

na(xa)n−1+nb(yb)n−1dydx=0.na(xa)n−1+nb(yb)n−1dydx=0.

⇒dydx=−na(xa)n−1bn(by)n−1=−(ba)n(xy)n−1.⇒dydx=−na(xa)n−1bn(by)n−1=−(ba)n(xy)n−1.

At the point (a,b),(a,b), the value of dydxdydx is −(ba)n(ab)n−1=−ba.−(ba)n(ab)n−1=−ba.

⇒⇒ The slope of the tangent to the curve at point (a,b)(a,b) is −ba.−ba.

So, the slope-point form of the equation of the tangent to the curve at point (a,b)(a,b) is

y−bx−a=−ba⇒a(y−b)=−b(x−a).y−bx−a=−ba⇒a(y−b)=−b(x−a).

⇒ay−ab=−bx+ab⇒bx+ay=2ab.⇒ay−ab=−bx+ab⇒bx+ay=2ab.

⇒xa+yb=2.⇒xa+yb=2.

⇒⇒ The equation of the tangent to the curve at point (a,b)(a,b) is xa+yb=2.